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Elena-2011 [213]

3 years ago

12

What number is the square root of 13 between

1 answer:

Semmy [17]3 years ago

8 0

Answer:

D

Step-by-step explanation:

3:

12

4:

16

Therefore 13 is in between 12 and 16

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Need help combining like terms

yaroslaw [1]

Answer:

-10x+7.5

Step-by-step explanation:

-2.5(4x-3)

Multiply each term by -2.5

-2.5 * 4x = -10x

-2.5 * 3 = 7.5

-10x + 7.5

7 0

3 years ago

An airplane started at point K, travelled 560 miles to point L, adjusted its route and travelled another 575 miles to point M. I

S_A_V [24]

Answer:

Step-by-step explanation:

Using the cosine theorem:

$c^{2} =a^{2} +b^{2} -abcos\beta$

$952^{2} =560^{2} +575^{2} -560*575cos\beta$

$\beta =cos^{-1} (\frac{560^{2}+575^{2}-952^{2} }{2*560*575} )$

The angle is 114.014 degrees if we substract 90 degrees it means that at point L the plane changed its angle by 24 degrees

4 0

3 years ago

Gina bought the small plastic boxes below. A rectangular prism with a length of 5 centimeters, width of 4 centimeters, and heigh

STALIN [3.7K]

Answer:the answer is The container on the left is made of 2 centimeters squared more plastic.

Step-by-step explanation:

i took the test

7 0

3 years ago

Read 2 more answers

The graph of a system of equations will intersect at more than 1 point.

Dafna1 [17]

Answer:

sometimes

Step-by-step explanation:

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3 years ago

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A closed, rectangular-faced box with a square base is to be constructed using only 36 m2 of material. What should the height h a

kkurt [141]

Answer:

$b=h=\sqrt{6}$ m

Step-by-step explanation:

Let

Bas length of box=b

Height of box=h

Material used in constructing of box=36 square m

We have to find the height h and base length b of the box to maximize the volume of box.

Surface area of box= $2b^2+4bh$

$2b^2+4bh=36$

$b^2+2bh=18$

$2bh=18-b^2$

$h=\frac{18-b^2}{2b}$

Volume of box, V= $b^2h$

Substitute the values

$V=b^2\times \frac{18-b^2}{2b}$

$V=\frac{1}{2}(18b-b^3)$

Differentiate w. r.t b

$\frac{dV}{db}=\frac{1}{2}(18-3b^2)$

$\frac{dV}{db}=0$

$\frac{1}{2}(18-3b^2)=0$

$\implies 18-3b^2=0$

$\implies 3b^2=18$

$b^2=6$

$b=\pm \sqrt{6}$

$b=\sqrt{6}$

The negative value of b is not possible because length cannot be negative.

Again differentiate w.r.t b

$\frac{d^2V}{db^2}=-3b$

At $b=\sqrt{6}$

$\frac{d^2V}{db^2}=-3\sqrt{6}$

Hence, the volume of box is maximum at $b=\sqrt{6}$ .

$h=\frac{18-(\sqrt{6})^2}{2\sqrt{6}}$

$h=\frac{18-6}{2\sqrt{6}}$

$h=\frac{12}{2\sqrt{6}}$

$h=\sqrt{6}$

$b=h=\sqrt{6}$ m

7 0

3 years ago